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A236411 Let p(k) denote the k-th prime; a(n) = smallest p(m) > p(n) such that the n-2 differences between [p(n), p(n+1), ..., p(2n-2)] are the same as the n-2 differences between [p(m), p(m+1), ..., p(m+n-2)]. 2

%I

%S 5,11,13,101,37,1277,1279,1616603,57405419,51448351,76623356077,

%T 115438255651991,433241801791933

%N Let p(k) denote the k-th prime; a(n) = smallest p(m) > p(n) such that the n-2 differences between [p(n), p(n+1), ..., p(2n-2)] are the same as the n-2 differences between [p(m), p(m+1), ..., p(m+n-2)].

%e n=5: We take the four primes [p(5)=11, 13, 17, 19], whose successive differences are 2, 4, 2. The next time we see this sequence of differences is at [101, 103, 107, 109], so a(5) = 101.

%t (* This program generates the first ten terms of the sequence. To generate more would require significantly greater computing resources *) dbp[n_]:=Differences[ Prime[ Range[ n,2n-2]]]; With[{prs=Prime[Range[ 3500000]]}, First/@ Flatten[ Table[Select[Partition[Drop[prs,n],n-1,1], Differences[#]==dbp[n]&,1],{n,2,11}],1]] (* _Harvey P. Dale_, Feb 05 2014 *)

%o (PARI) A236411 = n->{d=vector(n-2,i,prime(n+i)-prime(n));forprime(p=prime(n+1),,for(k=1,#d,isprime(p+d[k])||next(2));for(k=1,#d,p+d[k]==nextprime(p+if(k>1,d[k-1])+1))||next(2));return(p))} \\ The second k-loop would suffice, but the first makes it 5x faster. Yields a(10), a(11) in ca. 3 sec (i7, 1.9Ghz). - _M. F. Hasler_, Feb 05 2014

%Y See A073615 for a very similar sequence.

%K nonn,more

%O 2,1

%A _Don Reble_, Feb 05 2014

%E Edited by _N. J. A. Sloane_, Feb 05 2014

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Last modified July 19 03:54 EDT 2019. Contains 325144 sequences. (Running on oeis4.)